Question

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 24 ft/s.

(a) At what rate is his distance from second base decreasing when he is halfway to first base?

(b) At what rate is his distance from third base increasing at the same moment?

Short answer

(a) The rate of decrease of y is \( - 10.7\;{\rm{ft/s}}\).

(b) The distance between the third base is increasing at a rate of \(10.7\;{\rm{ft/s}}\).

Step-by-step solution

Step 1:Find an answer for part (a)

Let the distance traveled by the ball bex. Then the speed of the ball is given as,\(\frac{{{\rm{d}}x}}{{{\rm{d}}t}} = 24\)

The figure below represents the baseball diamond.

By the property of triangle:

\({y^2} = {\left( {90 - x} \right)^2} + {90^2}\)

Differentiate the equation.

\(\begin{aligned}2y\frac{{{\rm{d}}y}}{{{\rm{d}}t}} &= 2\left( {90 - x} \right)\left( { - \frac{{{\rm{d}}x}}{{{\rm{d}}t}}} \right)\\\frac{{{\rm{d}}y}}{{{\rm{d}}t}} &= \left( {\frac{{90 - x}}{y}} \right)\left( { - \frac{{{\rm{d}}x}}{{{\rm{d}}t}}} \right)\end{aligned}\)

For \(x = 45\), the value of \(y\) is:

\(\begin{aligned}y &= \sqrt {{x^2} + {{90}^2}} \\ &= \sqrt {{{45}^2} + {{90}^2}} \\ &= 45\sqrt 5 \end{aligned}\)

Substitute 45 for x, \(45\sqrt 5 \) for \(y\), and 24 for \(\frac{{{\rm{d}}x}}{{{\rm{d}}t}}\) in the equation \(\frac{{{\rm{d}}y}}{{{\rm{d}}t}} = \left( {\frac{{90 - x}}{y}} \right)\left( { - \frac{{{\rm{d}}x}}{{{\rm{d}}t}}} \right)\).

\(\begin{aligned}\frac{{{\rm{d}}y}}{{{\rm{d}}t}} &= \left( {\frac{{90 - 45}}{{45\sqrt 5 }}} \right)\left( { - 24} \right)\\ &= - \frac{{24}}{{\sqrt 5 }}\\ \approx - 10.7\end{aligned}\)

Thus, the rate of decrease of y is \( - 10.7\;{\rm{ft/s}}\).

Find an answer for part (b)

The relation for distance from the third base is:

\({z^2} = {x^2} + {90^2}\)

Differentiate the equation.

\(\begin{aligned}2z\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= 2x\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + 0\\\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= \frac{x}{z}\frac{{{\rm{d}}x}}{{{\rm{d}}t}}\end{aligned}\)

Substitute 45 for x, \(45\sqrt 5 \) for z, and 24 for \(\frac{{{\rm{d}}x}}{{{\rm{d}}t}}\) in the equation \(\frac{{{\rm{d}}z}}{{{\rm{d}}t}} = \frac{x}{z}\frac{{{\rm{d}}x}}{{{\rm{d}}t}}\).

\(\begin{aligned}\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= \frac{{45}}{{45\sqrt 5 }}\left( {24} \right)\\ &= \frac{{24}}{{\sqrt 5 }}\\ \approx 10.7\end{aligned}\)

Thus, the distance between the third base is increasing at a rate of \(10.7\;{\rm{ft/s}}\).